Question

(0,6у-1)-0,2(3у+1)<5у-4

Answer 1

$(0,6y-1)-0,2(3y+1) textless 5y-4 \ \ 0.6y-1-0,6y-0,2 textless 5y-4 \ \ -1,2 textless 5y-4 \ \ -5y textless -4+1,2 \ \ -5y textless -2,8 \ \ y textgreater 0,56$

x∈$(0,56; +infty});$